Question: $\dfrac{ 9p - 6q }{ -2 } = \dfrac{ -9p + r }{ 4 }$ Solve for $p$.
Multiply both sides by the left denominator. $\dfrac{ 9p - 6q }{ -{2} } = \dfrac{ -9p + r }{ 4 }$ $-{2} \cdot \dfrac{ 9p - 6q }{ -{2} } = -{2} \cdot \dfrac{ -9p + r }{ 4 }$ $9p - 6q = -{2} \cdot \dfrac { -9p + r }{ 4 }$ Multiply both sides by the right denominator. $9p - 6q = -2 \cdot \dfrac{ -9p + r }{ {4} }$ ${4} \cdot \left( 9p - 6q \right) = {4} \cdot -2 \cdot \dfrac{ -9p + r }{ {4} }$ ${4} \cdot \left( 9p - 6q \right) = -2 \cdot \left( -9p + r \right)$ Distribute both sides ${4} \cdot \left( 9p - 6q \right) = -{2} \cdot \left( -9p + r \right)$ ${36}p - {24}q = {18}p - {2}r$ Combine $p$ terms on the left. ${36p} - 24q = {18p} - 2r$ ${18p} - 24q = -2r$ Move the $q$ term to the right. $18p - {24q} = -2r$ $18p = -2r + {24q}$ Isolate $p$ by dividing both sides by its coefficient. ${18}p = -2r + 24q$ $p = \dfrac{ -2r + 24q }{ {18} }$ All of these terms are divisible by $2$ $p = \dfrac{ -{1}r + {12}q }{ {9} }$